WAEC SSCE - General Mathematics - 2016
Ajụjụ 1 Ripọtì
Find the lower quartile of the distribution illustrated by the cumulative frequency curve
Akọwa Nkọwa
To find the lower quartile, we need to identify the point on the cumulative frequency curve that corresponds to 25% of the total frequency. From the graph, we see that the total frequency is 40, and 25% of this is 10. The point on the curve that corresponds to a frequency of 10 is at a value of 19.0. Therefore, the lower quartile is 19.0. Note that the cumulative frequency curve shows the cumulative frequency of data values up to and including each data point on the horizontal axis. So, we can read off quartiles and other percentiles directly from the graph.
Ajụjụ 2 Ripọtì
The roots of a quadratic equation are \(\frac{4}{3}\) and -\(\frac{3}{7}\). Find the equation
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ
Akọwa Nkọwa
The gradient of a line is the measure of its steepness, or slope. To find the gradient of the line PQ, we need to use the formula: Gradient (m) = change in y / change in x We can calculate the change in y by subtracting the y-coordinate of point P from the y-coordinate of point Q: 8 - 2 = 6 Similarly, we can calculate the change in x by subtracting the x-coordinate of point P from the x-coordinate of point Q: 5 - 1 = 4 Therefore, the gradient of the line PQ is: Gradient (m) = change in y / change in x = 6 / 4 = 3/2 Therefore, the answer is option C: \(\frac{3}{2}\).
Ajụjụ 5 Ripọtì
The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of \(\Delta\) QNP = 6cm, calculate the area of the trapezium.
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue
Akọwa Nkọwa
When two balls are drawn from the bag with replacement, there are a total of $9\times9=81$ possible outcomes, since there are 9 balls in the bag and we are replacing each ball after drawing. To find the probability that the first ball is red and the second ball is blue, we can use the multiplication rule of probability. The probability that the first ball is red is $\frac{5}{9}$, since there are 5 red balls out of 9 total balls in the bag. After replacing the first ball, there are still 9 balls in the bag, but now 4 of them are blue. So the probability that the second ball is blue, given that the first ball was red, is $\frac{4}{9}$. Therefore, the probability that the first ball is red and the second ball is blue is: $$\frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$ Hence the answer is $\frac{20}{81}$.
Ajụjụ 7 Ripọtì
The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x
Akọwa Nkọwa
The formula for the volume of a pyramid is given as: V = (1/3) * base_area * height Let the length of the rectangle be x, then the base area of the pyramid is given as: base_area = x * 6 From the question, we are told that the volume of the pyramid is 90cm3 and its height is 15cm. Substituting into the formula for the volume of a pyramid, we have: 90 = (1/3) * (x * 6) * 15 Multiplying both sides by 3 gives: 270 = 90 * 6x Dividing both sides by 90 gives: 3 = 2x Therefore, x = 3/2 = 1.5 Hence, the value of x is 3. Answer option A, 3, is the correct answer.
Ajụjụ 8 Ripọtì
In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140o, find, correct to the nearest cm, the length of the chord MN.
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
Factorize; (2x + 3y)2 - (x - 4y)2
Ajụjụ 10 Ripọtì
\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age
Akọwa Nkọwa
To find the median, we need to first arrange the ages in order from lowest to highest. Then, we can determine which age lies in the middle of the list. Arranging the ages in order of increasing magnitude, we have: $$13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17$$ There are a total of 10+24+8+5+3 = 50 ages in the list, which is an even number. To find the median, we need to take the average of the two middle ages. The two middle ages are the 25th and 26th ages in the list, which are both 14. Therefore, the median age is: $$(14 + 14)/2 = 14$$ So the correct answer is 14.
Ajụjụ 11 Ripọtì
A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r
Akọwa Nkọwa
The volume of a sphere of radius r is given by the formula: \(\frac{4}{3}\pi r^3\). The volume of a cylinder of radius 3cm and height 4cm is given by the formula: \(\pi (3cm)^2(4cm) = 36\pi cm^3\). According to the question, the volume of the sphere is equal to the volume of the cylinder. Therefore: \[\frac{4}{3}\pi r^3 = 36\pi\] Dividing both sides by \(\frac{4}{3}\pi\), we get: \[r^3 = 27\] Taking the cube root of both sides, we get: \[r = 3\] Therefore, the value of r is 3cm. Answer is correct.
Ajụjụ 12 Ripọtì
Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined
Akọwa Nkọwa
The given expression will be undefined when its denominator is equal to zero since division by zero is undefined. So we need to find the values of y that make the denominator zero. \begin{align*} y^2 + 4y - 21 &= 0\\ (y+7)(y-3) &= 0 \end{align*} The denominator is equal to zero when either y+7=0 or y-3=0. Therefore, the expression is undefined when y=-7 or y=3. So, the answer is (c) 3, -7.
Ajụjụ 14 Ripọtì
If log2(3x - 1) = 5, find x.
Akọwa Nkọwa
We are given that log2(3x - 1) = 5. Using the definition of logarithms, we know that 25 = 32 is equal to the expression inside the logarithm. That is, 3x - 1 = 32 Adding 1 to both sides, we get 3x = 33 Dividing by 3, we get x = 11 Therefore, the value of x is 11. Answer: 11.
Ajụjụ 15 Ripọtì
If cos \(\theta\) = x and sin 60o = x + 0.5 0o < \(\theta\) < 90o, find, correct to the nearest degree, the value of \(\theta\)
Ajụjụ 16 Ripọtì
Express 1975 correct to 2 significant figures
Ajụjụ 18 Ripọtì
The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)
Akọwa Nkọwa
To make s the subject of the relation P = S + \(\frac{sm^2}{nr}\), we need to isolate s on one side of the equation. First, we can start by moving the \(\frac{sm^2}{nr}\) term to the other side of the equation by subtracting it from both sides: P - \(\frac{sm^2}{nr}\) = S Next, we can solve for s by multiplying both sides of the equation by \(\frac{nr}{m^2}\): s = \(\frac{nr}{m^2}\)(P - \(\frac{sm^2}{nr}\)) Simplifying the right-hand side, we get: s = \(\frac{nrp}{m^2}\) - \(\frac{s}{m}\) To isolate s, we can add \(\frac{s}{m}\) to both sides of the equation: s + \(\frac{sm}{m^2}\) = \(\frac{nrp}{m^2}\) Simplifying the left-hand side, we get: s(\(\frac{m + 1}{m^2}\)) = \(\frac{nrp}{m^2}\) Finally, we can solve for s by dividing both sides of the equation by \(\frac{m+1}{m^2}\): s = \(\frac{nrp}{nr + m^2}\) Therefore, the answer is s = \(\frac{nrp}{nr + m^2}\).
Ajụjụ 21 Ripọtì
Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2
Ajụjụ 22 Ripọtì
The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The curved surface area of a cylinder is given by the formula: 2\(\pi\)rh. Given that the cylinder is 5cm high and its curved surface area is 110cm2, we can write: 2\(\pi\)rh = 110, where h = 5 Substituting the value of \(\pi = \frac{22}{7}\) and h = 5, we get: 2 x \(\frac{22}{7}\) x r x 5 = 110 Simplifying this expression, we get: r = \(\frac{7}{2}\) r = 3.5cm (to one decimal place) Therefore, the radius of the cylinder is approximately 3.5cm. Hence, the answer is 3.5cm.
Ajụjụ 23 Ripọtì
In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
Simplify; \(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)
Akọwa Nkọwa
We can simplify this expression by using the properties of exponents and simplifying the numbers. First, we can write 27 as 33 and 81 as 34. Next, we can simplify the numerator by using the distributive property of exponents: \begin{align*} \frac{3^{n-1} \times 27^{n+1}}{81^n} &= \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}} \\ &= \frac{3^{n-1} \times 3^{3n+3}}{3^{4n}} \\ &= \frac{3^{4n-1}}{3^{4n}} \times 3^{3n+3} \\ &= 3^{-1} \times 3^{3n+3} \\ &= 3^{3n+2} \\ &= 3^2 \times 3^{3n} \\ &= 9 \times 3^n \\ \end{align*} Therefore, the simplified expression is 9 * 3n. So, the answer is 9.
Ajụjụ 25 Ripọtì
Simplify: (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)
Akọwa Nkọwa
To simplify the expression (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\), we first need to evaluate the subtraction inside the parentheses: \begin{align*} \frac{3}{4} - \frac{2}{3} &= \frac{9}{12} - \frac{8}{12} \\ &= \frac{1}{12} \end{align*} So now we have: \begin{align*} (\frac{3}{4} - \frac{2}{3})\times 1\frac{1}{5} &= \frac{1}{12} \times \frac{6}{5} \\ &= \frac{1 \times 6}{12 \times 5} \\ &= \frac{1}{10} \end{align*} Therefore, the answer is \(\frac{1}{10}\).
Ajụjụ 26 Ripọtì
A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
Akọwa Nkọwa
We can use the distance formula to find the length PQ, which is the distance between points P and Q on the line. The distance formula is: distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) where (x1, y1) = P and (x2, y2) = Q. Plugging in the values: distance = \(\sqrt{(5-1)^2 + (8-2)^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = \(2\sqrt{13}\) Therefore, the length PQ is \(2\sqrt{13}\). Answer: \(2\sqrt{13}\).
Ajụjụ 27 Ripọtì
In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m.
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
On a map, 1cm represent 5km. Find the area on the map that represents 100km2.
Akọwa Nkọwa
If 1cm represents 5km on the map, then x cm will represent 100km2 on the map. To find x, we can use the formula for area of a square, which is A = s2. In this case, we want to solve for s, where A = 100 and s represents the side length on the map in centimeters. So, s2 = A s2 = 100 s = √100 s = 10cm Therefore, 10cm on the map represents 100km2 on the ground. To find the area on the map that represents 100km2, we need to find the area of a square with a side length of 10cm. Area = s2 Area = 10cm x 10cm Area = 100cm2 So, the area on the map that represents 100km2 is 100cm2, which is equal to 1cm x 1cm, 4cm2, or.
Ajụjụ 29 Ripọtì
The relation y = x2 + 2x + k passes through the point (2,0). Find the value of k
Akọwa Nkọwa
We know that the relation y = x2 + 2x + k passes through the point (2,0), which means that when x=2, y=0. So, substituting these values in the relation, we get: 0 = 22 + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k Therefore, k = -8. Hence, the value of k is -8.
Ajụjụ 30 Ripọtì
\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
The table shows the ages of students in a club. How many students are in the club?
Akọwa Nkọwa
To find out how many students are in the club, we need to add up the frequencies in the table. So, Number of 13-year-olds = 10 Number of 14-year-olds = 24 Number of 15-year-olds = 8 Number of 16-year-olds = 5 Number of 17-year-olds = 3 Total number of students = 10 + 24 + 8 + 5 + 3 = 50 Therefore, there are 50 students in the club.
Ajụjụ 31 Ripọtì
The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?
Akọwa Nkọwa
The problem can be solved using the formula for direct variation: d = kt^2 where d is the distance, t is the time, and k is the constant of variation. We can solve for k using the given information that the stone falls 45cm in 3 seconds: 45 = k(3)^2 45 = 9k k = 5 Now that we know k, we can use the formula to find how far the stone will fall in 6 seconds: d = 5(6)^2 d = 5(36) d = 180cm Therefore, the answer is 180cm.
Ajụjụ 32 Ripọtì
An object is 6m away from the base of a mast. If the angle of depression of the object from the top of the mast is 50o, find, correct to 2 decimal places, the height of the mast.
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother?
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
What sum of money will amount to D10,400 in 5 years at 6% simple interest?
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
In the diagram, O is the centre of the circle, < QPS = 100o, < PSQ = 60o and < QSR. Calculate < SQR
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?
Akọwa Nkọwa
Ajụjụ 40 Ripọtì
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8...
Akọwa Nkọwa
The given sequence is the Fibonacci sequence, where the first two terms are 0 and 1, and each subsequent term is the sum of the two preceding it. Therefore, the next three terms are: - 13 (8 + 5) - 21 (13 + 8) - 34 (21 + 13) Hence, the answer is 13, 21, 34.
Ajụjụ 43 Ripọtì
The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range
Akọwa Nkọwa
The range of a set of data is the difference between the maximum and minimum values in the set. In this case, the minimum mark is 3 and the maximum mark is 16. Therefore, the range is 16 - 3 = 13. Hence, the answer is 13.
Ajụjụ 44 Ripọtì
Examine M' \(\cap\) N from the venn diagram
Ajụjụ 45 Ripọtì
If 20(mod 9) is equivalent to y(mod 6), find y.
Ajụjụ 46 Ripọtì
In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW
Akọwa Nkọwa
Using Pythagoras theorem, we can find that |UW| = 2cm. Next, we notice that \(\Delta\)UTW is an isosceles triangle since |TX| = |WX|. Therefore, o. Using the fact that the angles in a triangle add up to 180o, we can find that o - o - (60o + o. Therefore, we can substitute this into the previous equation to get: 90o + o - (60o + o = 2o, and since \(\Delta\)UTW is an isosceles triangle, we have that o. Hence, the answer is (C) 75o.
Ajụjụ 47 Ripọtì
Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\)
Akọwa Nkọwa
To simplify the expression \(\frac{2}{1-x} - \frac{1}{x}\), we need to first find a common denominator. The denominator of the first fraction is \((1-x)\) and the denominator of the second fraction is \(x\). The common denominator of these two fractions is \(x(1-x)\). Now, we need to rewrite each fraction with this common denominator. For the first fraction, we can multiply the numerator and denominator by \(x\), giving us \(\frac{2x}{x(1-x)}\). For the second fraction, we can multiply the numerator and denominator by \((1-x)\), giving us \(\frac{-(1-x)}{x(1-x)}\). Putting these two fractions together, we get: \[\frac{2x}{x(1-x)} - \frac{1-x}{x(1-x)} = \frac{2x - (1-x)}{x(1-x)} = \frac{3x-1}{x(1-x)}\] Therefore, the answer is \(\frac{3x-1}{x(1-x)}\).
Ajụjụ 48 Ripọtì
Which of following is a valid conclusion from the premise. "Nigeria footballers are good footballers"?
Akọwa Nkọwa
The valid conclusion from the premise "Nigeria footballers are good footballers" is "Joseph is a Nigerian footballer therefore he is a good footballer". This is because the premise establishes that Nigeria footballers are good, so anyone who is a Nigeria footballer can be inferred to be a good footballer. Therefore, Joseph, who is a Nigerian footballer, can be concluded to be a good footballer. The other options are not valid because they do not follow logically from the given premise.
Ajụjụ 49 Ripọtì
Simplify \(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)
Akọwa Nkọwa
First, let's simplify the numerator: \begin{align*} (p-r)^2 - r^2 &= (p^2 - 2pr + r^2) - r^2 \\ &= p^2 - 2pr \end{align*} Now, let's factor the denominator: \begin{align*} 2p^2 - 4pr &= 2p(p - 2r) \end{align*} Substituting these results, we get: \begin{align*} \frac{(p-r)^2 - r^2}{2p^2 - 4pr} &= \frac{p^2 - 2pr}{2p(p - 2r)} \\ &= \frac{p(p-2r)}{2p(p-2r)} \\ &= \frac{1}{2} \end{align*} Therefore, the answer is: \boxed{\frac{1}{2}}.
Ajụjụ 50 Ripọtì
(a) Copy and complete the table of values, correct to one decimal place, for the relation \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 3.0 | 1.6 | -2.0 | -3.6 | -3.0 | 2.0 |
(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \(y = 3\sin x + 2\cos x\) for \(0°\leq x \leq 360°\).
(c) Use the graph to solve :
(i) \(3\sin x + 2\cos x = 0\)
(ii) \(2 + 2\cos x + 3\sin x = 0\).
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
(a) 
In the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \(36 cm^{2}\). Find the value of |QR|.
(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.
Ajụjụ 52 Ripọtì
(a) Using the method of completing the square, solve, correct to 2 decimal places, \(\frac{x - 2}{4} = \frac{x + 2}{2x}\).
(b) 
In the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :
(i) < SQT ; (ii) < PQT.
Ajụjụ 53 Ripọtì
Using ruler and a pair of compasses only,
(a) construct :
(i) \(\Delta\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°.
(ii) locus, \(l_{1}\), of points equidistant from Y and Z.
(iii) locus, \(l_{2}\), of points parallel to XY through Z.
(b) Locate M, the point of intersection of \(l_{1}\) and \(l_{2}\).
(c) Measure < ZMY.
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
| Scores | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 25 | 30 | x | 28 | 40 | 32 |
The table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225;
(a) How many times was the die thrown?
(b) Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.
None
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.
(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all 3 means of transport. The number, x, of travellerswho travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train :
(i) represent this information in a Venn diagram;
(ii) calculate the
(1) value of x ; (2) number who travelled by at least two means.
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
(a) Without using Mathematical tables or calculators, evaluate \(\frac{0.09 \times 1.21}{3.3 \times 0.00025}\), leaving the answer in standard form (Scientific Notation).
(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned was GH¢1,200, find, correct to 3 significant figures, the interest rate per annum.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
The weight (in kg) of 50 contestants at a competition is as follows:
65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.
(a) Construct a frequenct table for the discrete data.
(b) Calculate, correct to 2 decimal places, the;
(i) mean ; (ii) standard deviation of the data.
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line \(2x + y = 6\).
(b) The operation \(\Delta\) is defined on the set T = {2, 3, 5, 7} by \(x \Delta y = (x + y + xy) mod 8\).
(i) Construct modulo 8 table for the operation \(\Delta\) on the set T.
(ii) Use the the table to find: (a) \(2 \Delta (5 \Delta 7)\) ; (b) \(2 \Delta n = 5 \Delta 7\).
Ajụjụ 59 Ripọtì
(a) Solve : \(7(x + 4) - \frac{2}{3}(x - 6) \leq 2[x - 3(x + 5)]\)
(b) A transport company has a total of 20 vehicles made up of tricycle and taxicabs. Each tricycle carries 2 passengers while each taxicab carries four passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?
Ajụjụ 60 Ripọtì
(a) 
In the diagram, < KLM = x, < LMK = y, < KJH = r and < KGF = 110°. If 2x = r = y, find the value of x.
(b) Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was N600.00 gretaer than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?
None
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
(a) 
In the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.
(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\(cm^{2}\), calculate its radius. [Take \(\pi = \frac{22}{7}\)].
Ajụjụ 62 Ripọtì
(a) If \(\frac{3p + 4q}{3p - 4q} = 2\), find \(p : q\).
(b) 
The diagram shows the cross section of a bridge with a semi-circular hollow in the middle. If the perimeter of the cross section is 34 cm, calculate the :
(i) length PQ; (ii) area of the cross section.
[Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
